Leetcode #2736: Maximum Sum Queries
In this guide, we solve Leetcode #2736 Maximum Sum Queries in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
You are given two 0-indexed integer arrays nums1 and nums2, each of length n, and a 1-indexed 2D array queries where queries[i] = [xi, yi]. For the ith query, find the maximum value of nums1[j] + nums2[j] among all indices j (0 <= j < n), where nums1[j] >= xi and nums2[j] >= yi, or -1 if there is no j satisfying the constraints.
Quick Facts
- Difficulty: Hard
- Premium: No
- Tags: Stack, Binary Indexed Tree, Segment Tree, Array, Binary Search, Sorting, Monotonic Stack
Intuition
The problem structure suggests a monotonic decision, which makes binary search a natural fit.
By halving the search space each step, we reach the answer efficiently.
Approach
Search either directly on a sorted array or on the answer space using a check function.
Each check is fast, and the logarithmic search keeps the overall runtime low.
Steps:
- Define the search bounds.
- Check the mid point condition.
- Narrow the bounds until convergence.
Example
Input: nums1 = [4,3,1,2], nums2 = [2,4,9,5], queries = [[4,1],[1,3],[2,5]]
Output: [6,10,7]
Explanation:
For the 1st query xi = 4 and yi = 1, we can select index j = 0 since nums1[j] >= 4 and nums2[j] >= 1. The sum nums1[j] + nums2[j] is 6, and we can show that 6 is the maximum we can obtain.
For the 2nd query xi = 1 and yi = 3, we can select index j = 2 since nums1[j] >= 1 and nums2[j] >= 3. The sum nums1[j] + nums2[j] is 10, and we can show that 10 is the maximum we can obtain.
For the 3rd query xi = 2 and yi = 5, we can select index j = 3 since nums1[j] >= 2 and nums2[j] >= 5. The sum nums1[j] + nums2[j] is 7, and we can show that 7 is the maximum we can obtain.
Therefore, we return [6,10,7].
Python Solution
class BinaryIndexedTree:
__slots__ = ["n", "c"]
def __init__(self, n: int):
self.n = n
self.c = [-1] * (n + 1)
def update(self, x: int, v: int):
while x <= self.n:
self.c[x] = max(self.c[x], v)
x += x & -x
def query(self, x: int) -> int:
mx = -1
while x:
mx = max(mx, self.c[x])
x -= x & -x
return mx
class Solution:
def maximumSumQueries(
self, nums1: List[int], nums2: List[int], queries: List[List[int]]
) -> List[int]:
nums = sorted(zip(nums1, nums2), key=lambda x: -x[0])
nums2.sort()
n, m = len(nums1), len(queries)
ans = [-1] * m
j = 0
tree = BinaryIndexedTree(n)
for i in sorted(range(m), key=lambda i: -queries[i][0]):
x, y = queries[i]
while j < n and nums[j][0] >= x:
k = n - bisect_left(nums2, nums[j][1])
tree.update(k, nums[j][0] + nums[j][1])
j += 1
k = n - bisect_left(nums2, y)
ans[i] = tree.query(k)
return ans
Complexity
The time complexity is , and the space complexity is . The space complexity is .
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.